solution to  equation $a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$ Does the equation
$$ a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$$
have a closed-form solution for $\theta$?  What about the case where $a^2 + b^2 = 1$?
 A: Yes, if this is acceptable as closed-form for you.
Using the formula, you can solve for $\theta$ and have an answer in terms of nothing more complicated than arcsine and square roots.
$$\theta=\arcsin(c/\sqrt{a^2+b^2})-\phi$$
I'll leave the determination of $\phi$ up to you as an exercise.
A: Given your use of signs I'll take it that $a,b \ge 0$. In which case, if $R = \sqrt{a^2+b^2}$ and $\alpha = \tan^{-1} \dfrac{b}{a}$ then
$$a \cos \theta - b\sin \theta = R\cos(\theta + \alpha)$$
and so if $\left| c \right| \le R$, we can solve the equation to get
$$\theta = \cos^{-1} \dfrac{c}{R} - \alpha = \boxed{\cos^{-1} \dfrac{c}{\sqrt{a^2+b^2}} - \tan^{-1} \dfrac{b}{a}}$$
Of course, this is just one solution; we can obtain the others by considering symmetries.
A: Using the identity $ sin(x)^2 + cos(x)^2=1 $, we have,
$$a \cdot\, \cos(\theta) - b \cdot \sin(\theta) = c \Rightarrow  a\sqrt{1-\sin^2(\theta)}= b \cdot \sin(\theta) + c $$ 
$$ \Rightarrow  a^2\,(1-\sin^2(\theta)) = ( b\sin(\theta) +c)^2 $$
$$ (a^2 + b^2)\sin(\theta)^2 + 2bc \sin(\theta) + (c^2-a^2)=0\,. $$
The above equation is quadratic in $\sin(\theta)$. Solving the equation in $\sin(\theta)$ yields,
$$ \sin(\theta) = \frac{ -2bc \pm \sqrt{4b^2c^2-4(a^2+b^2)(c^2-a^2)}}{2(a^2+b^2)}  $$
I believe you know how to find theta from here. Do not forget that $-1\leq \sin(\theta) \leq 1$.
